Table of contents | |
Oxidation Number | |
Important Rules for Determining Oxidation Number | |
Stock Notation | |
Solved Examples |
The oxidation number represents the charge a central metal atom maintains even after dissociating from all ligands attached to it. The oxidation state, or oxidation number, of an atom within a chemical compound explains the electron loss, indicating the degree of oxidation. It is a theoretical charge that an atom would carry if its bonds were entirely ionic.
Example: What is the oxidation state of sulphur in H2SO4 and H2SO5?
Solution: (i) H2SO4: Since there is no overall charge on the compound the oxidation states must cancel out. Although the oxidation state of hydrides is +1, there are two in the compound that must be accounted for. The same is true for oxygen; although the oxidation number of oxygen is -2, there are four oxygens present accounting for a total of -8. The +2 state contribution from the hydrides and -8 from the oxygens results in a -6 charge. The oxidation state on sulphur must be +6 for the molecule to be neutral.
Structure of Sulphuric Acid
(ii) H2SO5: the oxidation number of S is +6 in H2SO5 due to the presence of a peroxy linkage bond in the molecule.
Structure of Caro's Acid
Example: What is the Oxidation Number of Nitrogen in Ammonium Nitrate i.e. NH4NO3?
Solution: The steps to find the Oxidation Number of Nitrogen in Ammonium Nitrate are:
Therefore, the oxidation number of Nitrogen in ammonium nitrate is -3 and +5.
Example: Determine the oxidation number of each element in the following compounds:
(a) BaO2 (b) (NH4)2MoO4 (c)Na3Co(NO2)6 (d) CS2
Solution: (a) If oxygen in BaO2 had an oxidation number of -2, then barium would need to have an oxidation number of +4. However, elements in Group IIA, like barium, can't have a +4 oxidation state. This compound must be barium peroxide, [Ba2+][O22-]. Therefore, barium is +2 and oxygen is -1.
(b) In (NH4)2MoO4, the NH4+ ion has hydrogen at +1 and nitrogen at -3. With two NH4+ ions, the other part must be an MoO42- ion, where molybdenum is -6 and oxygen is -2.
(c) Sodium always has an oxidation state of +1 in its compounds. This compound contains the Co(NO2)63- complex ion, with six NO2- ions where nitrogen is +3 and oxygen is -2. The oxidation state of cobalt is thus +3.
(d) The most electronegative element in a compound always has a negative oxidation number. Since sulphur typically forms -2 ions, the oxidation number of sulphur in CS2 is -2, and carbon is +4.
Example: Calculate the oxidation number of nitrogen (N) in NO.
Solution: The oxidation state of oxygen(O) = -2
Consider the oxidation state of nitrogen(N) = x
Now, x-2 = 0 => x = +2
Hence, the oxidation number of N in NO is +2.
Example: Calculate the oxidation number of sulphur (S) in sulphuric acid:H2SO4
Solution: The oxidation state of oxygen(O) = -2
The oxidation state of hydrogen(H) = +1
Consider the oxidation state of sulphur (S) = X
Now,
2(+1)+X+(−22(+1)+X +(−2×4)
−2+X+8=0
X = +6
Hence, the oxidation number for sulphur in sulphuric acid is +6.
Example: Find oxidation number of Cr2 in the formula K2Cr2O7
Solution: K2= +2; O7= (-2 × 7); Cr2= 2 × x
Therefore, 2+ 2x-14=0. Thus x=+6 (Oxidation state of chromium)
Example: Find oxidation state of N in the formula NH4NO3
Solution: There are two different nitrogen atoms in the compound. Therefore we need to do the calculation separately
In NH+4, or X +4 (+1) = +1
Therefore x= -3 (Oxidation state in NH+4). The oxidation state of N in NO–3
y+3 × (-2)= -1
y= +6-1 =5
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1. What is an oxidation number? |
2. How is the oxidation number determined in a compound? |
3. Can an atom have multiple oxidation numbers? |
4. What is the significance of oxidation numbers in chemistry? |
5. How do oxidation numbers relate to redox reactions? |
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